y = x^.5
y' = .5 x^-.5
sketching this I am going to reverse slope equation
y' at tangent = m = (y-yo) /( x-xo) = .5x^-.5
y' = m = (y-0)/(x+16) = .5 x^-.5
x^.5 /(x+16) = .5 x^-.5
x/(x+16) = .5
x = .5 x + 8
.5 x = 8
x = 16
so the tangent hits the curve at x = 16
where y = 4 so at
(16,4)
so this line goes through
(-16,0) and (16,4)
You can take it from there I am sure.
Find an equation of the tangent line to the graph of the function f through the point (x0, y0) not on the graph. To find the point of tangency (x, y) on the graph of f, solve the following equation of f '(x).
f '(x) = y0 − y/x0 − x
f(x) = √x
(x0, y0) = (-16, 0)
y=?
1 answer